Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

Two sides of a triangle have lengths $a \text{ cm}$ and $b \text{ cm}$. If the triangle is to have the maximum area,then the length of the median from the vertex containing the sides $a$ and $b$ is

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are

If the vertices of a parallelogram are $(0, 0)$,$(1, 0)$,$(2, 2)$,and $(1, 2)$,then the angle between its diagonals is:

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$